//Given a string containing just the characters '(' and ')', find the length of 
//the longest valid (well-formed) parentheses substring. 
//
// 
// Example 1: 
//
// 
//Input: s = "(()"
//Output: 2
//Explanation: The longest valid parentheses substring is "()".
// 
//
// Example 2: 
//
// 
//Input: s = ")()())"
//Output: 4
//Explanation: The longest valid parentheses substring is "()()".
// 
//
// Example 3: 
//
// 
//Input: s = ""
//Output: 0
// 
//
// 
// Constraints: 
//
// 
// 0 <= s.length <= 3 * 104 
// s[i] is '(', or ')'. 
// 
// Related Topics 字符串 动态规划 
// 👍 1306 👎 0


package leetcode.editor.cn;

import java.util.Deque;
import java.util.LinkedList;

//Java：Longest Valid Parentheses
 class P32LongestValidParentheses{
    public static void main(String[] args) {
        Solution solution = new P32LongestValidParentheses().new Solution();
        // TO TEST
    }
    //leetcode submit region begin(Prohibit modification and deletion)
class Solution {
        //1.通过栈
        public int longestValidParenthesesStack(String s) {
            int maxans = 0;
            Deque<Integer> stack = new LinkedList<Integer>();
            stack.push(-1);
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) == '(') {
                    stack.push(i);
                } else {
                    stack.pop();
                    if (stack.isEmpty()) {
                        stack.push(i);
                    } else {
                        maxans = Math.max(maxans, i - stack.peek());
                    }
                }
            }
            return maxans;
        }
        //2.动态规划
        public int longestValidParenthesesDp(String s) {
            int maxans = 0;
            int[] dp = new int[s.length()];
            for (int i = 1; i < s.length(); i++) {
                if (s.charAt(i) == ')') {
                    if (s.charAt(i - 1) == '(') {
                        dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
                    } else if (i - dp[i - 1] > 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
                        dp[i] = dp[i - 1] + ((i - dp[i - 1]) >= 2 ? dp[i - dp[i - 1] - 2] : 0) + 2;
                    }
                    maxans = Math.max(maxans, dp[i]);
                }
            }
            return maxans;
        }
        public int longestValidParentheses(String s) {
            int left = 0, right = 0, maxlength = 0;
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) == '(') {
                    left++;
                } else {
                    right++;
                }
                if (left == right) {
                    maxlength = Math.max(maxlength, 2 * right);
                } else if (right > left) {
                    left = right = 0;
                }
            }
            left = right = 0;
            for (int i = s.length() - 1; i >= 0; i--) {
                if (s.charAt(i) == '(') {
                    left++;
                } else {
                    right++;
                }
                if (left == right) {
                    maxlength = Math.max(maxlength, 2 * left);
                } else if (left > right) {
                    left = right = 0;
                }
            }
            return maxlength;
        }


}
//leetcode submit region end(Prohibit modification and deletion)

}